Proving if $x^3$ is even, then $x$ is even. 
Theorem:
  If $x$ is a positive integer and  $x^3$ is even, then $x$ is even.

My Proof by Contrapositive:
I. Assuming that $x$ is odd, then I will show that $x^3$ is odd.
II. $x$ is odd, so $x$ can be expressed as $2k+1$
III. $(2k+1)^3$
IV. $2k(4k^2 + 6k + 3) +1$
V. Since $2k(4k^2 + 6k + 3) + 1$ is an integer, we can conclude $x^3$ is odd.
Did I go about this correctly? Should step V say "$2$ multiplied by an integer and added by $1$ is odd therefore we can conclude $x^3$ is odd"?
 A: Looks generally good. Some nitpicky things to point out:

*

*In I, you should reiterate the assumption that $x$ is a positive integer.

*In II, you introduced a new variable. You should state your assumption that $k$ is some integer.

*In III and IV, you have written down some expressions, but not commented on what they have to do with the proof. You should relate them together with equal signs:

Hence, it follows from I that:
  $$
x^3 = (2k + 1)^3 = 2k(4k^2+6k+3)+1
$$
  
*In V, the key point from the definition of an odd integer is not that $2k(4k^2+6k+3)+1$ is an integer, but that $k(4k^2+6k+3)$ is an integer. Otherwise, all integers would be odd.
  

A: The general idea is correct, but III and IV are no steps in a proof - they are sides of an equality.
In V you probably meant the right thing (Since $k(4k^2+6k+3) =: m$ is an integer, so $x^3 = 2m+1$ is an odd integer)
To correctly formalize your proof, write something like this:
(We will prove by contraposition...) Since $x$ is odd, there is a $k\in\mathbb Z$ such that $x = 2k+1$. Thus
$$x^3 = (2k+1)^3 = 2\underbrace{k(4k^2+6k+3)}_{\in\mathbb Z} + 1$$
So $x^3$ is odd. QED
A: Your proof is essentially correct (i.e. it is correct).  Although it's lacking certain levels of explanation.  First and foremost, you make the assumption (without proof or statement) that an integer is either odd or even.  From that, you assume that if an integer is not odd that is is even and, vice versa, if an integer not even then it is odd.
By making the above assumptions you can restate your hypothesis as: If $x^3$ is even then $x$ is even (assuming positive integers already).  The contrapositive then becomes if $x$ is not even then $x^3$ is not even.  You interpret this to mean that if $x$ is odd then $x^3$ is odd.
Now you assume (what must be assumed) from the beginning which is that $x$ is odd means that $x = 2\lambda_o + 1$ where $\lambda_o$ is an integer and that $y$ is even means that $y = 2\lambda_e$ where $\lambda_e$ is an integer.  So that now $x = 2\lambda_o + 1 \rightarrow x^3 = 2^3\lambda_o^3 + 24\lambda_o^2 + 6\lambda_o + 1 = 2\lambda_o\left(4\lambda_o^2 + 6\lambda_o + 3\right) + 1$.  This gives that $x^3 = 2\lambda_o' + 1$ where $\lambda_o' = \lambda_o\left(4\lambda_o^2 + 6\lambda_o + 3\right)$.  Therefore if $x$ is odd then $x^3$ is odd.
The reason your proof is correct is that it's correct to assume that if an integer is odd, that it is not even.  But without proof (stating what it means to be odd vs. even), the above doesn't work and the proof isn't valid.
Here is a quick counterexample which shows why your proof is insufficient (without proving an integer is always either odd or even and never both).  Prove the following: If $x^3 = 1\ (\text{mod } 13)$ then $x = 1\ (\text{mod } 13)$.  Your proof would be that $x$ does not equal $x = 1\ (\text{mod } 13)$, so for instance $x = 13\lambda_0$ and thus $x^3 = 13^3\lambda_0^3 = 13 (169\lambda_0^3)$  which, mod $13$, equals $0$ and is not $1$ therefore the original hypothesis is true.  Now this is clearly not true since $x = 3 = 3\ (\text{mod } 13)$ and $3^3 = 27 = 1\ (\text{mod } 13)$.
The reason the above reasoning is false is because it is not true that if $x \neq 1\ (\text{mod } 13)$ that $x = 0\ (\text{mod } 13)$--which was the assumption used to prove the (invalid) hypothesis.  Therefore without proving that an integer is always either odd or even, you cannot prove your statement.  But this is seriously nitpicking because most people understand that assumption--and you made it correctly.
